What is the integral of xsin(2ax) ?

The integral of xsin(2ax) is 1/(4a^2)(-2axcos(2ax)+sin(2ax))+C

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integral of xsin(2ax)

\int x sin (2 a x) d x

\frac{1}{4 a ^{2}} (- 2 a x cos (2 a x) + sin (2 a x)) + C

Solution steps

\int x sin (2 a x) d x

Apply u-substitution

: \int \frac{u sin ( u )}{4 a ^{2}} d u

= \int \frac{u sin ( u )}{4 a ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4 a ^{2}} \cdot \int u sin (u) d u

Apply Integration By Parts

: - u cos (u) - \int - cos (u) d u

= \frac{1}{4 a ^{2}} (- u cos (u) - \int - cos (u) d u)

\int - cos (u) d u = - sin (u)

= \frac{1}{4 a ^{2}} (- u cos (u) - (- sin (u)))

Substitute back

u = 2 a x

= \frac{1}{4 a ^{2}} (- 2 a x cos (2 a x) - (- sin (2 a x)))

Simplify

= \frac{1}{4 a ^{2}} (- 2 a x cos (2 a x) + sin (2 a x))

Add a constant to the solution

= \frac{1}{4 a ^{2}} (- 2 a x cos (2 a x) + sin (2 a x)) + C

y = (1 + 6 x^{2}) (x - x^{2})

n = 1 \sum \infty \frac{5}{n ^{4}}

x \to \infty lim (\frac{x}{3 x})

\frac{d ^{2}}{d x ^{2}} (\frac{1}{4 x ^{2} + 5 x})

\frac{d}{d x} (cos (7 - 2 x))

What is the integral of xsin(2ax) ?
The integral of xsin(2ax) is 1/(4a^2)(-2axcos(2ax)+sin(2ax))+C